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PP1 (Shorter) Quant Section 2 (Medium) Q13

Solving the Problem Method 1 (Algebra) We know that if a point is on both curves/lines, it must be where the lines/curves intersect. So, the y values would be the same, so we can equate them to figure out the x coordinate of the intersections: $y = x + 2 = x^2$ $x^2-(x+2) = 0$ $x^2-x-2=0$ Now, we can solve this using any of the standard methods, but this particular equation is pretty straightforward to factorize: $x^2-x-2 = (x+1)(x-2)$ So, either $x+1=0$ OR $x-2=0$, so the valid x coordinates would be $-1$ or $2$. Now, we can just substitute the x values back into the equations to get the y values: $y = x+2 = x^2 = -1+2 = (-1)^2 = 1$ $y = x+2 = x^2 = 2+2 = (2)^2 = 4$ So, the points that appear on both lines would be (-1, 1) and (2, 4). Method 2 (BackSolving) We can also check through each answer choice to see if the coordinates are valid for both equations. We can plug in the x coordinate into both equations and which ones give the correct y values for both: <table border="1" cellpadding="1" cellspacing="1" style="width:700px;"> <thead> <tr> <th scope="col">Point</th> <th scope="col">x value</th> <th scope="col">Expected y value</th> <th scope="col">Result of equation 1: x+2</th> <th scope="col">Result of equation 2: x^2</th> </tr> </thead> <tbody> <tr> <td>(-2, 0)</td> <td>-2</td> <td>0</td> <td>-2 + 2 = 0</td> <td>(-2)^2 = 4</td> </tr> <tr> <td>(-1, 1)</td> <td>-1</td> <td>1</td> <td>-1 + 2 = 1</td> <td>(-1)^2 = 1</td> </tr> <tr> <td>(0, 2)</td> <td>0</td> <td>2</td> <td>0 + 2 = 2</td> <td>(0)^2 = 0</td> </tr> <tr> <td>(1, 1)</td> <td>1</td> <td>1</td> <td>1 + 2 = 3</td> <td>(1)^2 = 1</td> </tr> <tr> <td>(2, 4)</td> <td>2</td> <td>4</td> <td>2 + 2 = 4</td> <td>(2)^2 = 4</td> </tr> </tbody> </table> We can see that both equations give the expected y value only for the points (-1, 1) and (2, 4) So, the correct answers are B and E.